Physics: Asquare loop (length along one side = 20 cm) rotates in a constant magnetic field which has a magnitude of 2.0 t. at an instant when the angle between the field and the normal to the plane of the loop is equal to 20° and increasing at the rate of 10°/s, what is the magnitude of the induced emf in the loop?

Asquare loop (length along one side = 20 cm) rotates

Ответ:

keshaayro1618

15.03.2022

Physics

0.0048 V

Explanation:

The side of the coil is

L = 20 cm = 0.20 m

So the area is

A=L^2 = (0.20)^2=0.04 m^2

The magnetic flux through the coil at any instant t is

$\Phi = BA cos (\omega t)$

where

B = 2.0 T is the magnetic flux density

A is the area of the coil

$\omega$ is the angular speed of the coil

t is the time

We know that the coil rotates at an angular speed of 10°/s, which converted into radians is

$\omega = \frac{10 \cdot 2\pi}{360}=0.175 rad/s$

The magnitude of the induced emf in the loop is equal to the derivative of the magnetic flux, so:

$\epsilon =- \frac{d\Phi}{dt}=-\frac{d}{dt}(BA cos(\omega t) = \omega B A sin(\omega t)$ (1)

We want to know the magnitude of the emf when the angle is $\theta=20^{\circ}$ , so substituting this value of the angle into $(\omega t)$ in eq.(1), we find:

$\epsilon = (0.175)(2.0)(0.04) sin (20^{\circ})=0.0048 V$

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